To be able to determine the quadratic character of an arbitrary number mod p (p an odd prime), we. The first (and most delicate) case concerns 2

Size: px
Start display at page:

Download "To be able to determine the quadratic character of an arbitrary number mod p (p an odd prime), we. The first (and most delicate) case concerns 2"

Transcription

1 Quadratic Reciprocity To be able to determine the quadratic character of an arbitrary number mod p (p an odd prime), we need to be able to evaluate q for any prime q. The first (and most delicate) case concerns 2. Proposition Let p be an odd prime. (1) p 1 (mod 8) 2 = 1 (2) p ±3 (mod 8) 2 = 1 p (3) p 1 (mod 8) 2 = 1 Proof (1) Let p = 8k +1. By F T, x 8k 1 (x 4k 1)(x 4k +1) 0 (mod p) has 8k solutions, whence by Lagrange s Theorem, each of the factor polynomials x 4k ±1 0 (mod p) must have 4k solutions. Since k must be positive, there must exist some integer a (necessarily prime to p) that satisfies a 4k +1 0 (mod p). But then (a 2k 1) 2 + 2(a k ) 2 a 4k +1 0 (mod p),

2 so (a 2k 1) 2 2(a k ) 2 (mod p). Multiplication (twice) by the inverse of a k yields {(a 2k 1)(a k ) 1 } 2 2 (mod p), whence 2 is a quadratic residue. (2) If p ±3 (mod 8) but 2 is a quadratic residue mod p, then there must be an integer a (0 < a < p) which satisfies the congruence x 2 2 (mod p). Indeed, we may assume that a is odd, else we can replace a with p a, which also satisfies this congruence and must have opposite parity from a. Then a 2 2 = pk for some odd number k. As pk = a 2 2 < a 2 < p 2, it follows that k < p. Now, for any prime factor q of k which is also smaller than p we must have a 2 2 (modq). If q ±3 (mod 8), then we can apply this same argument to find an even smaller prime for which 2 is a quadratic residue; by this process, eventually we must come to the smallest prime congruent to ±3 (mod 8) for which 2 is a quadratic residue. For simplicity, let us suppose it was p itself. Then every prime factor q of k referred to above is not congruent to ±3 (mod 8). So q ±1 (mod 8) for all such q, whence k ±1 (mod 8) as well. Therefore, a 2 2 pk ±3 ±1 ±3 (mod 8) a 2 5, 1 (mod 8). But this is impossible, since neither 5 nor 1 is a quadratic residue mod 8. This contradiction shows that 2 must be a quadratic nonresidue mod p.

3 (3) As in the argument in (2) above, assume that p 1 (mod 8) but that 2 is a quadratic residue mod p; then there must be an integer a (0 < a < p) which satisfies the congruence x 2 2 (mod p) and is odd (else we can replace it with p a). As before, a = pk for some odd k. As pk = a < (p 1) 2 2 < p 2, it follows that k < p. Thus, for any prime factor q of k which must be smaller than p we must have a 2 2 (modq). We may assume that p is the smallest prime factor of a of the form 8k 1, else we can transfer this argument to whichever prime does satisfy this property. So we have that every prime factor q of k referred to above is not congruent to 1(mod 8). If q 3 (mod 8), then a 2 2 (modq) 2 = 1, but q 1 q 1 (mod 4), so 2 = 1 2 = 1, q q q q contradicting (2) above. So it must be that q 1 or 3 (mod 8). But the product of primes of this type is also of this type, so k 1 or 3 (mod 8) and it follows that a = pk ( 1)k 1 or 3 (mod 8). But then a 2 3, 5 ±3 (mod 8), which is impossible since 3 and 3 are both quadratic nonresidues mod 8. //

4 Corollary If p is an odd prime, 2 = ( 1) ( p2 1)/8. Proof Left as an exercise. // There are a number of interesting consequences that flow from knowing the quadratic character of 2 mod p. Corollary There are infinitely many primes p 3 (mod8). Proof Suppose there are only finitely many, namely the primes p 1, p 2,, p k. Let N = (p 1 p 2 p k ) Then any prime q that divides N satisfies (p 1 p 2 p k ) (mod q), whence 2 is a quadratic residue mod q. It follows that q 1 or 3 (mod 8). If all the prime factors of N were of the form q 1 (mod 8), then N 1 (mod 8), but this would mean that (p 1 p 2 p k ) (mod 8) (p 1 p 2 p k ) 2 1 (mod 8), contradicting that 1 is a quadratic nonresidue mod 8. So at least one of the prime factors q of N, which is necessarily not one of p 1, p 2,, p k, must be congruent to 3 (mod 8), contradicting the fact that we had already listed all primes of this type among the p s. //

5 Corollary Let p be a Germain prime, i.e., a prime for which q = 2p + 1 is also prime (named after Sophie Germain, , a student of Gauss and Legendre). If also p > 3 and p 3 (mod4), then the Mersenne number M p = 2 p 1 is composite. Proof Since p = 4k + 3 for some k, q = 8k + 7, 2 is a quadratic residue mod q. By Euler s Criterion, 2 (q 1)/2 1 (mod q), so q 2 (q 1)/2 1 = 2 p 1 = M p. Since p > 3, we have 2 p 1 > p +1 (true for all integers p > 3, not just for primes), so 2 p > 2p + 2 M p > q. Thus, M p is composite. // Now let s consider the quadratic character of the prime 3 mod p. Proposition If p > 3 is prime, 3 p p (mod 3). Proof If p 1 (mod3), then p = 3k +1 for some k and so 4(x 3k 1) = (x k 1)(4x 2k + 4x k + 4) = (x k 1)((2x k +1) 2 + 3) Working mod p, we know that the polynomial on the left has 3k roots (F T) so the two factors on the

6 right have a full complement of roots as well. In particular, the congruence (2x k +1) 2 3 (mod p) has 2k solutions. Since 2x k +1 0 (mod p) can have at most k solutions, there must be a nonzero solution y 2x k +1 (mod p) to y 2 3 (mod p). On the other hand, if 3 = 1, then there is a nonzero solution to y 2 3 (mod p). As there are two solutions of opposite parity, we may take y to be the odd one. If y = 2z +1, then 4(z 2 + z +1) (2z +1) y (mod p) whence z 2 + z +1 0 (mod p) z (mod p). That is, z has order 3 mod p. But then 3 p 1, so p 1 (mod3). // Corollary 3 = 1 iff p ±1 (mod12). 3 Proof = 1 3 = 1 when 1 = 3. These p two symbols are both equal to 1 when p is congruent to 1 modulo 3 and 4, and both equal 1 when p is congruent to 1 modulo 3 and 4. //

An interesting class of problems of a computational nature ask for the standard residue of a power of a number, e.g.,

An interesting class of problems of a computational nature ask for the standard residue of a power of a number, e.g., Binary exponentiation An interesting class of problems of a computational nature ask for the standard residue of a power of a number, e.g., What are the last two digits of the number 2 284? In the absence

More information

6. Find an inverse of a modulo m for each of these pairs of relatively prime integers using the method

6. Find an inverse of a modulo m for each of these pairs of relatively prime integers using the method Exercises Exercises 1. Show that 15 is an inverse of 7 modulo 26. 2. Show that 937 is an inverse of 13 modulo 2436. 3. By inspection (as discussed prior to Example 1), find an inverse of 4 modulo 9. 4.

More information

SOLUTIONS TO PROBLEM SET 5. Section 9.1

SOLUTIONS TO PROBLEM SET 5. Section 9.1 SOLUTIONS TO PROBLEM SET 5 Section 9.1 Exercise 2. Recall that for (a, m) = 1 we have ord m a divides φ(m). a) We have φ(11) = 10 thus ord 11 3 {1, 2, 5, 10}. We check 3 1 3 (mod 11), 3 2 9 (mod 11), 3

More information

Collection of rules, techniques and theorems for solving polynomial congruences 11 April 2012 at 22:02

Collection of rules, techniques and theorems for solving polynomial congruences 11 April 2012 at 22:02 Collection of rules, techniques and theorems for solving polynomial congruences 11 April 2012 at 22:02 Public Polynomial congruences come up constantly, even when one is dealing with much deeper problems

More information

UNIVERSITY OF MANITOBA DATE: December 7, FINAL EXAMINATION TITLE PAGE TIME: 3 hours EXAMINER: M. Davidson

UNIVERSITY OF MANITOBA DATE: December 7, FINAL EXAMINATION TITLE PAGE TIME: 3 hours EXAMINER: M. Davidson TITLE PAGE FAMILY NAME: (Print in ink) GIVEN NAME(S): (Print in ink) STUDENT NUMBER: SEAT NUMBER: SIGNATURE: (in ink) (I understand that cheating is a serious offense) INSTRUCTIONS TO STUDENTS: This is

More information

LECTURE 3: CONGRUENCES. 1. Basic properties of congruences We begin by introducing some definitions and elementary properties.

LECTURE 3: CONGRUENCES. 1. Basic properties of congruences We begin by introducing some definitions and elementary properties. LECTURE 3: CONGRUENCES 1. Basic properties of congruences We begin by introducing some definitions and elementary properties. Definition 1.1. Suppose that a, b Z and m N. We say that a is congruent to

More information

NUMBER THEORY AMIN WITNO

NUMBER THEORY AMIN WITNO NUMBER THEORY AMIN WITNO.. w w w. w i t n o. c o m Number Theory Outlines and Problem Sets Amin Witno Preface These notes are mere outlines for the course Math 313 given at Philadelphia

More information

MATH 324 Elementary Number Theory Solutions to Practice Problems for Final Examination Monday August 8, 2005

MATH 324 Elementary Number Theory Solutions to Practice Problems for Final Examination Monday August 8, 2005 MATH 324 Elementary Number Theory Solutions to Practice Problems for Final Examination Monday August 8, 2005 Deartment of Mathematical and Statistical Sciences University of Alberta Question 1. Find integers

More information

Math 127: Equivalence Relations

Math 127: Equivalence Relations Math 127: Equivalence Relations Mary Radcliffe 1 Equivalence Relations Relations can take many forms in mathematics. In these notes, we focus especially on equivalence relations, but there are many other

More information

Discrete Square Root. Çetin Kaya Koç Winter / 11

Discrete Square Root. Çetin Kaya Koç  Winter / 11 Discrete Square Root Çetin Kaya Koç koc@cs.ucsb.edu Çetin Kaya Koç http://koclab.cs.ucsb.edu Winter 2017 1 / 11 Discrete Square Root Problem The discrete square root problem is defined as the computation

More information

LECTURE 7: POLYNOMIAL CONGRUENCES TO PRIME POWER MODULI

LECTURE 7: POLYNOMIAL CONGRUENCES TO PRIME POWER MODULI LECTURE 7: POLYNOMIAL CONGRUENCES TO PRIME POWER MODULI 1. Hensel Lemma for nonsingular solutions Although there is no analogue of Lagrange s Theorem for prime power moduli, there is an algorithm for determining

More information

Foundations of Cryptography

Foundations of Cryptography Foundations of Cryptography Ville Junnila viljun@utu.fi Department of Mathematics and Statistics University of Turku 2015 Ville Junnila viljun@utu.fi Lecture 10 1 of 17 The order of a number (mod n) Definition

More information

Quadratic Residues. Legendre symbols provide a computational tool for determining whether a quadratic congruence has a solution. = a (p 1)/2 (mod p).

Quadratic Residues. Legendre symbols provide a computational tool for determining whether a quadratic congruence has a solution. = a (p 1)/2 (mod p). Quadratic Residues 4--015 a is a quadratic residue mod m if x = a (mod m). Otherwise, a is a quadratic nonresidue. Quadratic Recirocity relates the solvability of the congruence x = (mod q) to the solvability

More information

Is 1 a Square Modulo p? Is 2?

Is 1 a Square Modulo p? Is 2? Chater 21 Is 1 a Square Modulo? Is 2? In the revious chater we took various rimes and looked at the a s that were quadratic residues and the a s that were nonresidues. For examle, we made a table of squares

More information

MAT Modular arithmetic and number theory. Modular arithmetic

MAT Modular arithmetic and number theory. Modular arithmetic Modular arithmetic 1 Modular arithmetic may seem like a new and strange concept at first The aim of these notes is to describe it in several different ways, in the hope that you will find at least one

More information

b) Find all positive integers smaller than 200 which leave remainder 1, 3, 4 upon division by 3, 5, 7 respectively.

b) Find all positive integers smaller than 200 which leave remainder 1, 3, 4 upon division by 3, 5, 7 respectively. Solutions to Exam 1 Problem 1. a) State Fermat s Little Theorem and Euler s Theorem. b) Let m, n be relatively prime positive integers. Prove that m φ(n) + n φ(m) 1 (mod mn). Solution: a) Fermat s Little

More information

Number Theory - Divisibility Number Theory - Congruences. Number Theory. June 23, Number Theory

Number Theory - Divisibility Number Theory - Congruences. Number Theory. June 23, Number Theory - Divisibility - Congruences June 23, 2014 Primes - Divisibility - Congruences Definition A positive integer p is prime if p 2 and its only positive factors are itself and 1. Otherwise, if p 2, then p

More information

MTH 3527 Number Theory Quiz 10 (Some problems that might be on the quiz and some solutions.) 1. Euler φ-function. Desribe all integers n such that:

MTH 3527 Number Theory Quiz 10 (Some problems that might be on the quiz and some solutions.) 1. Euler φ-function. Desribe all integers n such that: MTH 7 Number Theory Quiz 10 (Some roblems that might be on the quiz and some solutions.) 1. Euler φ-function. Desribe all integers n such that: (a) φ(n) = Solution: n = 4,, 6 since φ( ) = ( 1) =, φ() =

More information

Solutions to Problem Set 6 - Fall 2008 Due Tuesday, Oct. 21 at 1:00

Solutions to Problem Set 6 - Fall 2008 Due Tuesday, Oct. 21 at 1:00 18.781 Solutions to Problem Set 6 - Fall 008 Due Tuesday, Oct. 1 at 1:00 1. (Niven.8.7) If p 3 is prime, how many solutions are there to x p 1 1 (mod p)? How many solutions are there to x p 1 (mod p)?

More information

The Chinese Remainder Theorem

The Chinese Remainder Theorem The Chinese Remainder Theorem Theorem. Let n 1,..., n r be r positive integers relatively prime in pairs. (That is, gcd(n i, n j ) = 1 whenever 1 i < j r.) Let a 1,..., a r be any r integers. Then the

More information

#A3 INTEGERS 17 (2017) A NEW CONSTRAINT ON PERFECT CUBOIDS. Thomas A. Plick

#A3 INTEGERS 17 (2017) A NEW CONSTRAINT ON PERFECT CUBOIDS. Thomas A. Plick #A3 INTEGERS 17 (2017) A NEW CONSTRAINT ON PERFECT CUBOIDS Thomas A. Plick tomplick@gmail.com Received: 10/5/14, Revised: 9/17/16, Accepted: 1/23/17, Published: 2/13/17 Abstract We show that out of the

More information

SOLUTIONS FOR PROBLEM SET 4

SOLUTIONS FOR PROBLEM SET 4 SOLUTIONS FOR PROBLEM SET 4 A. A certain integer a gives a remainder of 1 when divided by 2. What can you say about the remainder that a gives when divided by 8? SOLUTION. Let r be the remainder that a

More information

Solutions to Exam 1. Problem 1. a) State Fermat s Little Theorem and Euler s Theorem. b) Let m, n be relatively prime positive integers.

Solutions to Exam 1. Problem 1. a) State Fermat s Little Theorem and Euler s Theorem. b) Let m, n be relatively prime positive integers. Solutions to Exam 1 Problem 1. a) State Fermat s Little Theorem and Euler s Theorem. b) Let m, n be relatively rime ositive integers. Prove that m φ(n) + n φ(m) 1 (mod mn). c) Find the remainder of 1 008

More information

Modular Arithmetic. Kieran Cooney - February 18, 2016

Modular Arithmetic. Kieran Cooney - February 18, 2016 Modular Arithmetic Kieran Cooney - kieran.cooney@hotmail.com February 18, 2016 Sums and products in modular arithmetic Almost all of elementary number theory follows from one very basic theorem: Theorem.

More information

Carmen s Core Concepts (Math 135)

Carmen s Core Concepts (Math 135) Carmen s Core Concepts (Math 135) Carmen Bruni University of Waterloo Week 7 1 Congruence Definition 2 Congruence is an Equivalence Relation (CER) 3 Properties of Congruence (PC) 4 Example 5 Congruences

More information

The congruence relation has many similarities to equality. The following theorem says that congruence, like equality, is an equivalence relation.

The congruence relation has many similarities to equality. The following theorem says that congruence, like equality, is an equivalence relation. Congruences A congruence is a statement about divisibility. It is a notation that simplifies reasoning about divisibility. It suggests proofs by its analogy to equations. Congruences are familiar to us

More information

Solutions for the Practice Questions

Solutions for the Practice Questions Solutions for the Practice Questions Question 1. Find all solutions to the congruence 13x 12 (mod 35). Also, answer the following questions about the solutions to the above congruence. Are there solutions

More information

Primitive Roots. Chapter Orders and Primitive Roots

Primitive Roots. Chapter Orders and Primitive Roots Chapter 5 Primitive Roots The name primitive root applies to a number a whose powers can be used to represent a reduced residue system modulo n. Primitive roots are therefore generators in that sense,

More information

Wilson s Theorem and Fermat s Theorem

Wilson s Theorem and Fermat s Theorem Wilson s Theorem and Fermat s Theorem 7-27-2006 Wilson s theorem says that p is prime if and only if (p 1)! = 1 (mod p). Fermat s theorem says that if p is prime and p a, then a p 1 = 1 (mod p). Wilson

More information

Assignment 2. Due: Monday Oct. 15, :59pm

Assignment 2. Due: Monday Oct. 15, :59pm Introduction To Discrete Math Due: Monday Oct. 15, 2012. 11:59pm Assignment 2 Instructor: Mohamed Omar Math 6a For all problems on assignments, you are allowed to use the textbook, class notes, and other

More information

1.6 Congruence Modulo m

1.6 Congruence Modulo m 1.6 Congruence Modulo m 47 5. Let a, b 2 N and p be a prime. Prove for all natural numbers n 1, if p n (ab) and p - a, then p n b. 6. In the proof of Theorem 1.5.6 it was stated that if n is a prime number

More information

An elementary study of Goldbach Conjecture

An elementary study of Goldbach Conjecture An elementary study of Goldbach Conjecture Denise Chemla 26/5/2012 Goldbach Conjecture (7 th, june 1742) states that every even natural integer greater than 4 is the sum of two odd prime numbers. If we

More information

Applications of Fermat s Little Theorem and Congruences

Applications of Fermat s Little Theorem and Congruences Applications of Fermat s Little Theorem and Congruences Definition: Let m be a positive integer. Then integers a and b are congruent modulo m, denoted by a b mod m, if m (a b). Example: 3 1 mod 2, 6 4

More information

Discrete Math Class 4 ( )

Discrete Math Class 4 ( ) Discrete Math 37110 - Class 4 (2016-10-06) 41 Division vs congruences Instructor: László Babai Notes taken by Jacob Burroughs Revised by instructor DO 41 If m ab and gcd(a, m) = 1, then m b DO 42 If gcd(a,

More information

Calculators will not be permitted on the exam. The numbers on the exam will be suitable for calculating by hand.

Calculators will not be permitted on the exam. The numbers on the exam will be suitable for calculating by hand. Midterm #2: practice MATH 311 Intro to Number Theory midterm: Thursday, Oct 20 Please print your name: Calculators will not be permitted on the exam. The numbers on the exam will be suitable for calculating

More information

Introduction to Number Theory 2. c Eli Biham - November 5, Introduction to Number Theory 2 (12)

Introduction to Number Theory 2. c Eli Biham - November 5, Introduction to Number Theory 2 (12) Introduction to Number Theory c Eli Biham - November 5, 006 345 Introduction to Number Theory (1) Quadratic Residues Definition: The numbers 0, 1,,...,(n 1) mod n, are called uadratic residues modulo n.

More information

PT. Primarity Tests Given an natural number n, we want to determine if n is a prime number.

PT. Primarity Tests Given an natural number n, we want to determine if n is a prime number. PT. Primarity Tests Given an natural number n, we want to determine if n is a prime number. (PT.1) If a number m of the form m = 2 n 1, where n N, is a Mersenne number. If a Mersenne number m is also a

More information

Exam 1 7 = = 49 2 ( ) = = 7 ( ) =

Exam 1 7 = = 49 2 ( ) = = 7 ( ) = Exam 1 Problem 1. a) Define gcd(a, b). Using Euclid s algorithm comute gcd(889, 168). Then find x, y Z such that gcd(889, 168) = x 889 + y 168 (check your answer!). b) Let a be an integer. Prove that gcd(3a

More information

The Chinese Remainder Theorem

The Chinese Remainder Theorem The Chinese Remainder Theorem Theorem. Let m and n be two relatively prime positive integers. Let a and b be any two integers. Then the two congruences x a (mod m) x b (mod n) have common solutions. Any

More information

University of British Columbia. Math 312, Midterm, 6th of June 2017

University of British Columbia. Math 312, Midterm, 6th of June 2017 University of British Columbia Math 312, Midterm, 6th of June 2017 Name (please be legible) Signature Student number Duration: 90 minutes INSTRUCTIONS This test has 7 problems for a total of 100 points.

More information

Math 255 Spring 2017 Solving x 2 a (mod n)

Math 255 Spring 2017 Solving x 2 a (mod n) Math 255 Spring 2017 Solving x 2 a (mod n) Contents 1 Lifting 1 2 Solving x 2 a (mod p k ) for p odd 3 3 Solving x 2 a (mod 2 k ) 5 4 Solving x 2 a (mod n) for general n 9 1 Lifting Definition 1.1. Let

More information

Practice Midterm 2 Solutions

Practice Midterm 2 Solutions Practice Midterm 2 Solutions May 30, 2013 (1) We want to show that for any odd integer a coprime to 7, a 3 is congruent to 1 or 1 mod 7. In fact, we don t need the assumption that a is odd. By Fermat s

More information

ON THE EQUATION a x x (mod b) Jam Germain

ON THE EQUATION a x x (mod b) Jam Germain ON THE EQUATION a (mod b) Jam Germain Abstract. Recently Jimenez and Yebra [3] constructed, for any given a and b, solutions to the title equation. Moreover they showed how these can be lifted to higher

More information

Modular Arithmetic. claserken. July 2016

Modular Arithmetic. claserken. July 2016 Modular Arithmetic claserken July 2016 Contents 1 Introduction 2 2 Modular Arithmetic 2 2.1 Modular Arithmetic Terminology.................. 2 2.2 Properties of Modular Arithmetic.................. 2 2.3

More information

CHAPTER 2. Modular Arithmetic

CHAPTER 2. Modular Arithmetic CHAPTER 2 Modular Arithmetic In studying the integers we have seen that is useful to write a = qb + r. Often we can solve problems by considering only the remainder, r. This throws away some of the information,

More information

The Strong Finiteness of Double Mersenne Primes and the Infinity of Root Mersenne Primes and Near-square Primes of Mersenne Primes

The Strong Finiteness of Double Mersenne Primes and the Infinity of Root Mersenne Primes and Near-square Primes of Mersenne Primes The Strong Finiteness of Double Mersenne Primes and the Infinity of Root Mersenne Primes and Near-square Primes of Mersenne Primes Pingyuan Zhou E-mail:zhoupingyuan49@hotmail.com Abstract In this paper

More information

Distribution of Primes

Distribution of Primes Distribution of Primes Definition. For positive real numbers x, let π(x) be the number of prime numbers less than or equal to x. For example, π(1) = 0, π(10) = 4 and π(100) = 25. To use some ciphers, we

More information

PRIMES IN SHIFTED SUMS OF LUCAS SEQUENCES. Lenny Jones Department of Mathematics, Shippensburg University, Shippensburg, Pennsylvania

PRIMES IN SHIFTED SUMS OF LUCAS SEQUENCES. Lenny Jones Department of Mathematics, Shippensburg University, Shippensburg, Pennsylvania #A52 INTEGERS 17 (2017) PRIMES IN SHIFTED SUMS OF LUCAS SEQUENCES Lenny Jones Department of Mathematics, Shippensburg University, Shippensburg, Pennsylvania lkjone@ship.edu Lawrence Somer Department of

More information

ALGEBRA: Chapter I: QUESTION BANK

ALGEBRA: Chapter I: QUESTION BANK 1 ALGEBRA: Chapter I: QUESTION BANK Elements of Number Theory Congruence One mark questions: 1 Define divisibility 2 If a b then prove that a kb k Z 3 If a b b c then PT a/c 4 If a b are two non zero integers

More information

Constructions of Coverings of the Integers: Exploring an Erdős Problem

Constructions of Coverings of the Integers: Exploring an Erdős Problem Constructions of Coverings of the Integers: Exploring an Erdős Problem Kelly Bickel, Michael Firrisa, Juan Ortiz, and Kristen Pueschel August 20, 2008 Abstract In this paper, we study necessary conditions

More information

Goldbach Conjecture (7 th june 1742)

Goldbach Conjecture (7 th june 1742) Goldbach Conjecture (7 th june 1742) We note P the odd prime numbers set. P = {p 1 = 3, p 2 = 5, p 3 = 7, p 4 = 11,...} n 2N\{0, 2, 4}, p P, p n/2, q P, q n/2, n = p + q We call n s Goldbach decomposition

More information

Solutions for the Practice Final

Solutions for the Practice Final Solutions for the Practice Final 1. Ian and Nai play the game of todo, where at each stage one of them flips a coin and then rolls a die. The person who played gets as many points as the number rolled

More information

A REMARK ON A PAPER OF LUCA AND WALSH 1. Zhao-Jun Li Department of Mathematics, Anhui Normal University, Wuhu, China. Min Tang 2.

A REMARK ON A PAPER OF LUCA AND WALSH 1. Zhao-Jun Li Department of Mathematics, Anhui Normal University, Wuhu, China. Min Tang 2. #A40 INTEGERS 11 (2011) A REMARK ON A PAPER OF LUCA AND WALSH 1 Zhao-Jun Li Department of Mathematics, Anhui Normal University, Wuhu, China Min Tang 2 Department of Mathematics, Anhui Normal University,

More information

Number Theory. Konkreetne Matemaatika

Number Theory. Konkreetne Matemaatika ITT9131 Number Theory Konkreetne Matemaatika Chapter Four Divisibility Primes Prime examples Factorial Factors Relative primality `MOD': the Congruence Relation Independent Residues Additional Applications

More information

BAND SURGERY ON KNOTS AND LINKS, III

BAND SURGERY ON KNOTS AND LINKS, III BAND SURGERY ON KNOTS AND LINKS, III TAIZO KANENOBU Abstract. We give two criteria of links concerning a band surgery: The first one is a condition on the determinants of links which are related by a band

More information

ELEMENTS OF NUMBER THEORY & CONGRUENCES. Lagrange, Legendre and Gauss. Mth Mathematicst

ELEMENTS OF NUMBER THEORY & CONGRUENCES. Lagrange, Legendre and Gauss. Mth Mathematicst ELEMENTS OF NUMBER THEORY & CONGRUENCES Lagrange, Legendre and Gauss ELEMENTS OF NUMBER THEORY & CONGRUENCES 1) If a 0, b 0 Z and a/b, b/a then 1) a=b 2) a=1 3) b=1 4) a=±b Ans : is 4 known result. If

More information

MA/CSSE 473 Day 9. The algorithm (modified) N 1

MA/CSSE 473 Day 9. The algorithm (modified) N 1 MA/CSSE 473 Day 9 Primality Testing Encryption Intro The algorithm (modified) To test N for primality Pick positive integers a 1, a 2,, a k < N at random For each a i, check for a N 1 i 1 (mod N) Use the

More information

MT 430 Intro to Number Theory MIDTERM 2 PRACTICE

MT 430 Intro to Number Theory MIDTERM 2 PRACTICE MT 40 Intro to Number Theory MIDTERM 2 PRACTICE Material covered Midterm 2 is comrehensive but will focus on the material of all the lectures from February 9 u to Aril 4 Please review the following toics

More information

5 Symmetric and alternating groups

5 Symmetric and alternating groups MTHM024/MTH714U Group Theory Notes 5 Autumn 2011 5 Symmetric and alternating groups In this section we examine the alternating groups A n (which are simple for n 5), prove that A 5 is the unique simple

More information

Sheet 1: Introduction to prime numbers.

Sheet 1: Introduction to prime numbers. Option A Hand in at least one question from at least three sheets Sheet 1: Introduction to prime numbers. [provisional date for handing in: class 2.] 1. Use Sieve of Eratosthenes to find all prime numbers

More information

CMPSCI 250: Introduction to Computation. Lecture #14: The Chinese Remainder Theorem David Mix Barrington 4 October 2013

CMPSCI 250: Introduction to Computation. Lecture #14: The Chinese Remainder Theorem David Mix Barrington 4 October 2013 CMPSCI 250: Introduction to Computation Lecture #14: The Chinese Remainder Theorem David Mix Barrington 4 October 2013 The Chinese Remainder Theorem Infinitely Many Primes Reviewing Inverses and the Inverse

More information

Cryptography CS 555. Topic 20: Other Public Key Encryption Schemes. CS555 Topic 20 1

Cryptography CS 555. Topic 20: Other Public Key Encryption Schemes. CS555 Topic 20 1 Cryptography CS 555 Topic 20: Other Public Key Encryption Schemes Topic 20 1 Outline and Readings Outline Quadratic Residue Rabin encryption Goldwasser-Micali Commutative encryption Homomorphic encryption

More information

Chapter 4 Cyclotomic Cosets, the Mattson Solomon Polynomial, Idempotents and Cyclic Codes

Chapter 4 Cyclotomic Cosets, the Mattson Solomon Polynomial, Idempotents and Cyclic Codes Chapter 4 Cyclotomic Cosets, the Mattson Solomon Polynomial, Idempotents and Cyclic Codes 4.1 Introduction Much of the pioneering research on cyclic codes was carried out by Prange [5]inthe 1950s and considerably

More information

The covering congruences of Paul Erdős. Carl Pomerance Dartmouth College

The covering congruences of Paul Erdős. Carl Pomerance Dartmouth College The covering congruences of Paul Erdős Carl Pomerance Dartmouth College Conjecture (Erdős, 1950): For each number B, one can cover Z with finitely many congruences to distinct moduli all > B. Erdős (1995):

More information

Math 412: Number Theory Lecture 6: congruence system and

Math 412: Number Theory Lecture 6: congruence system and Math 412: Number Theory Lecture 6: congruence system and classes Gexin Yu gyu@wm.edu College of William and Mary Chinese Remainder Theorem Chinese Remainder Theorem: let m 1, m 2,..., m k be pairwise coprimes.

More information

Zhanjiang , People s Republic of China

Zhanjiang , People s Republic of China Math. Comp. 78(2009), no. 267, 1853 1866. COVERS OF THE INTEGERS WITH ODD MODULI AND THEIR APPLICATIONS TO THE FORMS x m 2 n AND x 2 F 3n /2 Ke-Jian Wu 1 and Zhi-Wei Sun 2, 1 Department of Mathematics,

More information

Introduction to Modular Arithmetic

Introduction to Modular Arithmetic 1 Integers modulo n 1.1 Preliminaries Introduction to Modular Arithmetic Definition 1.1.1 (Equivalence relation). Let R be a relation on the set A. Recall that a relation R is a subset of the cartesian

More information

arxiv:math/ v2 [math.ho] 23 Aug 2008

arxiv:math/ v2 [math.ho] 23 Aug 2008 arxiv:math/0507401v2 [math.ho] 23 Aug 2008 An easy method for finding many very large prime numbers Leonhard Euler 1. It suffices to explain this method with a single example of numbers contained in the

More information

FORBIDDEN INTEGER RATIOS OF CONSECUTIVE POWER SUMS

FORBIDDEN INTEGER RATIOS OF CONSECUTIVE POWER SUMS FORBIDDEN INTEGER RATIOS OF CONSECUTIVE POWER SUMS IOULIA N. BAOULINA AND PIETER MOREE To the memory of Prof. Wolfgang Schwarz Abstract. Let S k (m) := 1 k + 2 k +... + (m 1) k denote a power sum. In 2011

More information

Fermat s little theorem. RSA.

Fermat s little theorem. RSA. .. Computing large numbers modulo n (a) In modulo arithmetic, you can always reduce a large number to its remainder a a rem n (mod n). (b) Addition, subtraction, and multiplication preserve congruence:

More information

Number Theory/Cryptography (part 1 of CSC 282)

Number Theory/Cryptography (part 1 of CSC 282) Number Theory/Cryptography (part 1 of CSC 282) http://www.cs.rochester.edu/~stefanko/teaching/11cs282 1 Schedule The homework is due Sep 8 Graded homework will be available at noon Sep 9, noon. EXAM #1

More information

Fall. Spring. Possible Summer Topics

Fall. Spring. Possible Summer Topics Fall Paper folding: equilateral triangle (parallel postulate and proofs of theorems that result, similar triangles), Trisect a square paper Divisibility by 2-11 and by combinations of relatively prime

More information

by Michael Filaseta University of South Carolina

by Michael Filaseta University of South Carolina by Michael Filaseta University of South Carolina Background: A covering of the integers is a system of congruences x a j (mod m j, j =, 2,..., r, with a j and m j integral and with m j, such that every

More information

Power = 36² mod 99 Power = 9 5 a 5 = 0 x = 81 Power = 9² mod 99 Power = 81 6 a 6 = 1 x = 81 x 81 mod 99 x = 27 7 a 7 = 1 x = 27 x 27 mod 99 x = 36

Power = 36² mod 99 Power = 9 5 a 5 = 0 x = 81 Power = 9² mod 99 Power = 81 6 a 6 = 1 x = 81 x 81 mod 99 x = 27 7 a 7 = 1 x = 27 x 27 mod 99 x = 36 Question 1 Section 4.1 11. What time does a 12-hour clock read a) 80 hours after it reads 11:00? b) 40 hours before it reads 12:00? c) 100 hours after it reads 6:00? I don't really understand this question

More information

Congruence. Solving linear congruences. A linear congruence is an expression in the form. ax b (modm)

Congruence. Solving linear congruences. A linear congruence is an expression in the form. ax b (modm) Congruence Solving linear congruences A linear congruence is an expression in the form ax b (modm) a, b integers, m a positive integer, x an integer variable. x is a solution if it makes the congruence

More information

A theorem on the cores of partitions

A theorem on the cores of partitions A theorem on the cores of partitions Jørn B. Olsson Department of Mathematical Sciences, University of Copenhagen Universitetsparken 5,DK-2100 Copenhagen Ø, Denmark August 9, 2008 Abstract: If s and t

More information

Permutation Groups. Every permutation can be written as a product of disjoint cycles. This factorization is unique up to the order of the factors.

Permutation Groups. Every permutation can be written as a product of disjoint cycles. This factorization is unique up to the order of the factors. Permutation Groups 5-9-2013 A permutation of a set X is a bijective function σ : X X The set of permutations S X of a set X forms a group under function composition The group of permutations of {1,2,,n}

More information

Multiples and Divisibility

Multiples and Divisibility Multiples and Divisibility A multiple of a number is a product of that number and an integer. Divisibility: A number b is said to be divisible by another number a if b is a multiple of a. 45 is divisible

More information

Two congruences involving 4-cores

Two congruences involving 4-cores Two congruences involving 4-cores ABSTRACT. The goal of this paper is to prove two new congruences involving 4- cores using elementary techniques; namely, if a 4 (n) denotes the number of 4-cores of n,

More information

A Study of Relationship Among Goldbach Conjecture, Twin prime and Fibonacci number

A Study of Relationship Among Goldbach Conjecture, Twin prime and Fibonacci number A Study of Relationship Among Goldbach Conjecture, Twin and Fibonacci number Chenglian Liu Department of Computer Science, Huizhou University, China chenglianliu@gmailcom May 4, 015 Version 48 1 Abstract

More information

Implementation / Programming: Random Number Generation

Implementation / Programming: Random Number Generation Introduction to Modeling and Simulation Implementation / Programming: Random Number Generation OSMAN BALCI Professor Department of Computer Science Virginia Polytechnic Institute and State University (Virginia

More information

MATH 433 Applied Algebra Lecture 12: Sign of a permutation (continued). Abstract groups.

MATH 433 Applied Algebra Lecture 12: Sign of a permutation (continued). Abstract groups. MATH 433 Applied Algebra Lecture 12: Sign of a permutation (continued). Abstract groups. Permutations Let X be a finite set. A permutation of X is a bijection from X to itself. The set of all permutations

More information

x 8 (mod 15) x 8 3 (mod 5) eli 2 2y 6 (mod 10) y 3 (mod 5) 6x 9 (mod 11) y 3 (mod 11) So y = 3z + 3u + 3w (mod 990) z = (990/9) (990/9) 1

x 8 (mod 15) x 8 3 (mod 5) eli 2 2y 6 (mod 10) y 3 (mod 5) 6x 9 (mod 11) y 3 (mod 11) So y = 3z + 3u + 3w (mod 990) z = (990/9) (990/9) 1 Exercise help set 6/2011 Number Theory 1. x 2 0 (mod 2) x 2 (mod 6) x 2 (mod 3) a) x 5 (mod 7) x 5 (mod 7) x 8 (mod 15) x 8 3 (mod 5) (x 8 2 (mod 3)) So x 0y + 2z + 5w + 8u (mod 210). y is not needed.

More information

The Sign of a Permutation Matt Baker

The Sign of a Permutation Matt Baker The Sign of a Permutation Matt Baker Let σ be a permutation of {1, 2,, n}, ie, a one-to-one and onto function from {1, 2,, n} to itself We will define what it means for σ to be even or odd, and then discuss

More information

TOPOLOGY, LIMITS OF COMPLEX NUMBERS. Contents 1. Topology and limits of complex numbers 1

TOPOLOGY, LIMITS OF COMPLEX NUMBERS. Contents 1. Topology and limits of complex numbers 1 TOPOLOGY, LIMITS OF COMPLEX NUMBERS Contents 1. Topology and limits of complex numbers 1 1. Topology and limits of complex numbers Since we will be doing calculus on complex numbers, not only do we need

More information

LUCAS-SIERPIŃSKI AND LUCAS-RIESEL NUMBERS

LUCAS-SIERPIŃSKI AND LUCAS-RIESEL NUMBERS LUCAS-SIERPIŃSKI AND LUCAS-RIESEL NUMBERS DANIEL BACZKOWSKI, OLAOLU FASORANTI, AND CARRIE E. FINCH Abstract. In this paper, we show that there are infinitely many Sierpiński numbers in the sequence of

More information

Final exam. Question Points Score. Total: 150

Final exam. Question Points Score. Total: 150 MATH 11200/20 Final exam DECEMBER 9, 2016 ALAN CHANG Please present your solutions clearly and in an organized way Answer the questions in the space provided on the question sheets If you run out of room

More information

CMPSCI 250: Introduction to Computation. Lecture #14: The Chinese Remainder Theorem David Mix Barrington 24 February 2012

CMPSCI 250: Introduction to Computation. Lecture #14: The Chinese Remainder Theorem David Mix Barrington 24 February 2012 CMPSCI 250: Introduction to Computation Lecture #14: The Chinese Remainder Theorem David Mix Barrington 24 February 2012 The Chinese Remainder Theorem Infinitely Many Primes Reviewing Inverses and the

More information

ON MODULI FOR WHICH THE FIBONACCI SEQUENCE CONTAINS A COMPLETE SYSTEM OF RESIDUES S. A. BURR Belt Telephone Laboratories, Inc., Whippany, New Jersey

ON MODULI FOR WHICH THE FIBONACCI SEQUENCE CONTAINS A COMPLETE SYSTEM OF RESIDUES S. A. BURR Belt Telephone Laboratories, Inc., Whippany, New Jersey ON MODULI FOR WHICH THE FIBONACCI SEQUENCE CONTAINS A COMPLETE SYSTEM OF RESIDUES S. A. BURR Belt Telephone Laboratories, Inc., Whippany, New Jersey Shah [1] and Bruckner [2] have considered the problem

More information

MATHEMATICS ON THE CHESSBOARD

MATHEMATICS ON THE CHESSBOARD MATHEMATICS ON THE CHESSBOARD Problem 1. Consider a 8 8 chessboard and remove two diametrically opposite corner unit squares. Is it possible to cover (without overlapping) the remaining 62 unit squares

More information

Solutions for the 2nd Practice Midterm

Solutions for the 2nd Practice Midterm Solutions for the 2nd Practice Midterm 1. (a) Use the Euclidean Algorithm to find the greatest common divisor of 44 and 17. The Euclidean Algorithm yields: 44 = 2 17 + 10 17 = 1 10 + 7 10 = 1 7 + 3 7 =

More information

Cryptography, Number Theory, and RSA

Cryptography, Number Theory, and RSA Cryptography, Number Theory, and RSA Joan Boyar, IMADA, University of Southern Denmark November 2015 Outline Symmetric key cryptography Public key cryptography Introduction to number theory RSA Modular

More information

Math 319 Problem Set #7 Solution 18 April 2002

Math 319 Problem Set #7 Solution 18 April 2002 Math 319 Problem Set #7 Solution 18 April 2002 1. ( 2.4, problem 9) Show that if x 2 1 (mod m) and x / ±1 (mod m) then 1 < (x 1, m) < m and 1 < (x + 1, m) < m. Proof: From x 2 1 (mod m) we get m (x 2 1).

More information

arxiv: v3 [math.co] 4 Dec 2018 MICHAEL CORY

arxiv: v3 [math.co] 4 Dec 2018 MICHAEL CORY CYCLIC PERMUTATIONS AVOIDING PAIRS OF PATTERNS OF LENGTH THREE arxiv:1805.05196v3 [math.co] 4 Dec 2018 MIKLÓS BÓNA MICHAEL CORY Abstract. We enumerate cyclic permutations avoiding two patterns of length

More information

Diffie-Hellman key-exchange protocol

Diffie-Hellman key-exchange protocol Diffie-Hellman key-exchange protocol This protocol allows two users to choose a common secret key, for DES or AES, say, while communicating over an insecure channel (with eavesdroppers). The two users

More information

Degree project NUMBER OF PERIODIC POINTS OF CONGRUENTIAL MONOMIAL DYNAMICAL SYSTEMS

Degree project NUMBER OF PERIODIC POINTS OF CONGRUENTIAL MONOMIAL DYNAMICAL SYSTEMS Degree project NUMBER OF PERIODIC POINTS OF CONGRUENTIAL MONOMIAL DYNAMICAL SYSTEMS Author: MD.HASIRUL ISLAM NAZIR BASHIR Supervisor: MARCUS NILSSON Date: 2012-06-15 Subject: Mathematics and Modeling Level:

More information

Outline Introduction Big Problems that Brun s Sieve Attacks Conclusions. Brun s Sieve. Joe Fields. November 8, 2007

Outline Introduction Big Problems that Brun s Sieve Attacks Conclusions. Brun s Sieve. Joe Fields. November 8, 2007 Big Problems that Attacks November 8, 2007 Big Problems that Attacks The Sieve of Eratosthenes The Chinese Remainder Theorem picture Big Problems that Attacks Big Problems that Attacks Eratosthene s Sieve

More information

p 1 MAX(a,b) + MIN(a,b) = a+b n m means that m is a an integer multiple of n. Greatest Common Divisor: We say that n divides m.

p 1 MAX(a,b) + MIN(a,b) = a+b n m means that m is a an integer multiple of n. Greatest Common Divisor: We say that n divides m. Great Theoretical Ideas In Computer Science Steven Rudich CS - Spring Lecture Feb, Carnegie Mellon University Modular Arithmetic and the RSA Cryptosystem p- p MAX(a,b) + MIN(a,b) = a+b n m means that m

More information

Countability. Jason Filippou UMCP. Jason Filippou UMCP) Countability / 12

Countability. Jason Filippou UMCP. Jason Filippou UMCP) Countability / 12 Countability Jason Filippou CMSC250 @ UMCP 06-23-2016 Jason Filippou (CMSC250 @ UMCP) Countability 06-23-2016 1 / 12 Outline 1 Infinity 2 Countability of integers and rationals 3 Uncountability of R Jason

More information

The tenure game. The tenure game. Winning strategies for the tenure game. Winning condition for the tenure game

The tenure game. The tenure game. Winning strategies for the tenure game. Winning condition for the tenure game The tenure game The tenure game is played by two players Alice and Bob. Initially, finitely many tokens are placed at positions that are nonzero natural numbers. Then Alice and Bob alternate in their moves

More information